Question

Suppose Player 1 sentences Player 2 to be hanged in one the next T days (T is an integer). Moreover, P1 tells P2 that the day of hanging will be a surprise. Formally, there are T possible periods in the game. In any of the rest T-1 periods of the game, Player 1 chooses whether to hang (H) or not (N), and Player 2 simultaneously chooses (i.e. guesses) whether he will be hanged (h) or not (n). If Player 1 chooses H in any of the rest T-1 periods, the game ends at the end of that period. Otherwise, the game proceeds to the next period. [Note: at the end of each period, players.actions in that period become publicly observable.] In the Tth period, Player 1 has only one action available: H (player 2 still has his two actions). Note that there is a terminal node only after a period in which Player 1 chooses H. Payoffs at any terminal node are (-1; 1) if player 2 chose h in that period (hence, he was not surprised) and (1;-1) if player 2 chose n in that period (hence, he was surprised).

(hint: apply backward induction to find the Nash equilibrium first)

Q1: Suppose T = 2: Is there a Nash equilibrium that is not subgame perfect?

Q2: Suppose T = 3: Is there a Nash equilibrium that is not subgame perfect?

Answer 1

When considering the question in the context of game theory, focusing on the Nash equilibrium and subgame perfect equilibrium, it appears that for both scenarios with T = 2 and T = 3, the Nash equilibria that could form would be subgame perfect due to the paradox of the surprise hanging.

Step-by-step explanation:

The scenario provided is a theoretical problem in game theory, more specifically involving Nash equilibrium and subgame perfect equilibrium, which is related to the concept of strategic decision-making. Backward induction is the process of reasoning backwards from the end of a problem or scenario to determine a sequence of optimal actions. This concept is used in both game theory and decision making to ensure the outcome of a game or a decision is the best possible for a player given the actions of other players.

In the context of the question, if T = 2 or T = 3, we aim to determine whether a Nash equilibrium that is not subgame perfect exists. For T = 2, by backward induction, the conclusion would be that Player 2 can predict a hanging on the first day (since it wouldn't be a surprise on the last). Thus, Player 1 has no winning strategy that would be a surprise. No pure strategy Nash equilibrium exists here, so they are all subgame perfect. For T = 3, the paradox unwinds one more layer, but similar logic applies, and any Nash equilibrium would also be subgame perfect as Player 2 can always anticipate the hanging based on the logic of elimination.